13. x11+2x1 +26x = 600 cos 10t; x(0) = 10, x1(0) = 0

To solve these problems, we need to find both the steady periodic solution xsp(t)x_{\text{sp}}(t)xsp​(t) and the actual solution x(t)x(t)x(t), which is a combination of the steady periodic solution and a transient term xtr(t)x_{\text{tr}}(t)xtr​(t). Here's the general approach for each problem:

1. Steady Periodic Solution xsp(t)x_{\text{sp}}(t)xsp​(t)

The given differential equations are likely non-homogeneous, and the steady periodic solution corresponds to the particular solution that oscillates at the driving frequency. It will be of the form:

xsp(t)=Ccos⁡(ωt−α)x_{\text{sp}}(t) = C \cos(\omega t - \alpha)xsp​(t)=Ccos(ωt−α)

To find CCC and α\alphaα:

• Convert the forcing function into a form with sinusoidal terms.
• Use the method of undetermined coefficients or phasor analysis to match the amplitude and phase with the forcing function.
• This will give you the steady-state response to the external forcing term.

2. Transient Solution xtr(t)x_{\text{tr}}(t)xtr​(t)

The transient solution xtr(t)x_{\text{tr}}(t)xtr​(t) solves the homogeneous equation. Typically, the general solution to the homogeneous part is:

xtr(t)=Aeλ1t+Beλ2tx_{\text{tr}}(t) = A e^{\lambda_1 t} + B e^{\lambda_2 t}xtr​(t)=Aeλ1​t+Beλ2​t

where λ1\lambda_1λ1​ and λ2\lambda_2λ2​ are the characteristic roots. For second-order linear equations, these roots depend on the damping and can represent:

• Overdamping: real and distinct roots,
• Critical damping: real and repeated roots,
• Underdamping: complex conjugate roots.

3. Actual Solution x(t)x(t)x(t)

The actual solution is the sum of the steady-state and transient solutions:

x(t)=xsp(t)+xtr(t)x(t) = x_{\text{sp}}(t) + x_{\text{tr}}(t)x(t)=xsp​(t)+xtr​(t)

4. Applying Initial Conditions

To determine the constants AAA and BBB, apply the given initial conditions x(0)x(0)x(0) and x′(0)x'(0)x′(0) to the actual solution x(t)x(t)x(t) and solve for the unknowns.

 13. x"+2x'+26x=600cos10t; x(0)=10, x'(0)=0